91Ó°ÊÓ

Two parallel non-conducting rings with their central axes along a common line:

• Ring 1 has a uniform charge $q_{\mathit{1}}$ and radius R.
• Ring 2 has a uniform charge role="math" localid="1657361420698" $q_{\mathit{2}}$with radius R.
• The separation distance, d=3R.
• The net electric field on the common line at point P, which is at a distance of R is zero.

Using the concept of the electric field at an axial pint, we can find the net electric field of the individual rings. Further simplifying the equation, we will get the ratio of the charges.

Formula:

The magnitude of the electric field at an axial point,

$E=\frac{q}{4{\mathrm{Ï€Î\mu }}_0{\left({\mathrm{d}}^2+R^{\mathit{2}}\right)}^{3/2}}$ (i)

Where d is the distance of field point from the charge, R is the radius of the circular ring, q is charge of the particle.

Weuse the expressionof electric field, assuming both charges are positive at point P using equation (i) as follows: (net electric at R from ring 1 is zero.)

$E_{leftring}=E_{rightring}$

$$\begin{gathered} \frac{1}{4{\mathrm{Ï€Î\mu }}_0}\frac{q_{1}R}{{\left(R^2+R^2\right)}^{{\displaystyle \frac{3}{2}}}}=\frac{1}{4\mathrm{Ï€Î\mu }}\frac{q_2\left(2R\right)}{({\left(2R{)}^2+R^2\right)}^{3/2}} \\ \frac{q_1}{q_2}=2{\left(\frac{2}{5}\right)}^{3/2} \\ ≈0.0506 \end{gathered}$$

Hence, the value of the required ratio is 0.506.